How do you solve #1/3abs(1/2x-2) >= 2#?

1 Answer
May 22, 2018

Answer:

#(-oo,-8]uu[16,oo)#

Explanation:

#"inequalities of the type "|x|>=a#

#"always have solutions of the form"#

#x<= -a" or "x>=a#

#"isolate "|1/2x-2|" by dividing both sides by "1/3#

#rArr|1/2-2|>=6#

#rArr1/2x-2<=-6" or "1/2x-2>=6#

#"add 2 to both sides of both inequalities"#

#rArr1/2x<=-4" or "1/2x>=8#

#"multiply both sides of both inequalities by 2"#

#rArrx <=-8" or "x>=16#

#x in(-oo,-8]uu[16,oo)larrcolor(blue)"in interval notation"#